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Question
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
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Solution
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(bb(cos^2θ) + bb(sin^2θ))/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ...............[sin2θ + cos2θ = 1]
= `1/sinθ xx 1/bbcosθ`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
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